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SOLUTIONS 1 a F (A,B,C) = A' B' C' A' B' C A B' C' A B' C A B C' A B C Distributive = A'B' (C' C) AB' (C' C) AB (CDec 16, 08 · Both f (1) and f (2)=2 There is c in (0,2) such that f' (c)=f (b)f (a)/ba By the MVT, there is a number c in (0, 1) such that f' (c) = f (1) f (0)/ (1 0) = 2/1 = 2 You can use the same idea to show that for another number c in (1, 2), f' (c) = 0 If you knew that f' was continuous, you could use the Intermediate Value Theorem to showDec 17, 18 · 3 Answers3 Consider the set of x, call it S, where x < a You seek for the probability, P ( S) Any expression that lead to S produces the exact same probability, b, irrespective of its decleration If f ( x) is a monotonic (strictly) increasing function, x < a directly implies f ( x) < f ( a) and vice versa, ie if f ( x) < f ( a), then x < a
F(b) = y = exp( b2=2 C) = Kexp( b2=2) But f(b) is a pdf, and hence the only constant that Kcan be is the one that makes this integrate to 1 K= p 1 2ˇ Therefore f(x) = p 2ˇ e x2=2, the pdf of a standard normal distribution 4 (424) Let X 1;;X n be iid random varaibles with common pdf with f(x) = e x for x> , 0 elsewhere, where 1CD/DVD f B A ABluray f B X N AVHS e v f B A ̔ A R s A v X ̊ Ѓu j h A CD/DVD f B A ABluray f B X N AVHS e v f B A ̔ A R s A v X ̊ Brand New DoorAvanti F&B, a collective eatery A Denver, CO
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Nov 24, 17 · a=1, b=1 If a function f(x) has an inverse function f^1(x), color(red)(f(f^1(x))=x) is always true Then, f(f^1(x))=x ⇔ a(bxa) b =x ⇔ abxa^2b=x It must be an identity for x Hence, ab=1 (1) and a^2b=0 (2) are satisfied From the equation (2), b=a^2 Substitute this into (1), a^3=1 a^3=1 a=1 (assuming that a is a real number) b=(1)^2=11 day ago · Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack ExchangeJul 07, 13 · P X X ^ f B @ a @ ɍs @ ̂R 800 P X X ^ f B V Y ǂ݂ɂȂ Ă F l A 肪 Ƃ ܂ B ҂̂ Q ł B
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1222 (a) Prove that f(A ∩ B) = f(A) ∩ f(B) for all A,B ⊆ X iff f is injective Proof We show the implications separately =⇒ Let x 1,x 2 ∈ X be arbitrary with f(x 1) = f(x 2) Let A = {x 1} and B = {x 2} By assumption, f(A∩B) = f(A)∩f(B) = {f(x 1)}∩{f(x 2)} = {f(x 1)} This implies that there exists an element x ∈ A ∩F(x) hasa value equal to f(c) = ∫b a f(x)dx b a Multiplying bothsides by b a proves the result 4The first fundamental theorem of integral calculus We are now in a position to prove our first major result about the definite integral The result concerns the socalled area function F(x) = ∫ x a f(t)dt and its derivative with respect to xQ5 (26(10)) Let p be an odd prime, and suppose that a 6 0(mod p) Show that if the congruence x2 a(mod pj) has a solution when j= 1, then it has a solution for all j Proof Suppose b(mod p) is a solution We claim the derivative of the polynomial x2 a(= 2x) does not vanish at b(mod p)
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Adjoining finitely many algebraic elements (ie the roots of p(x)), we know that E is a finite extension of Z p Therefore, E is isomorphic to GF(pn) for some n Since every element of GF(pn) is a root of the polynomial xpn − x, the same is true of the elements of E Since E contains the roots of p(x), the roots of p(x) must also be roots{ f R A e B X g ́A f R A g ̕ y Ɛ Z p ̔ W A ړI Ƃ A f R A g Ɋ֘A Ƃ𒆐S Ƃ Đݗ ꂽ c ̂ł B u f R A e B X g v Ƃ́A f R d E f R O b Y ɑ \ 郉 C X g Ȃǂ g p f R V Z @ g f R A g h ̐ Z p ҂ w ܂ BX is rational, and f(x) = 1 if x is irrational Let P be any partition of 0,1 Since the jth subinterval contains both rational and irrational numbers, we have mj = 0 and Mj = 1 Therefore L(f;P) = j=1 0(xj − xj−1) = 0 and U(f;P) = j=1 1(xj − xj−1) = 1 so the lower integral is 0 and the upper integral is 1 Therefore f is not
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